Question

So from a question asked in another thread, I have thought of a new question and the answer is not obvious to me.

So it appears there is a c++ rule that says if you have a const reference to a temporary, then the lifetime of the temporary is at least as long as the const reference. But what if you have a local const reference to another object’s member variable and then when you leave scope – Does it call the destructor of that variable?

So here is modified program from the original question:

#include <iostream>
#include <string>
using namespace std;

class A {
public:
   A(std::string l) { k = l; };
   std::string get() const { return k; };
   std::string k;
};

class B {
public:
   B(A a) : a(a) {}
   void b() { cout << a.get(); }  //Has a member function
   A a;
};

void f(const A& a)
{  //Gets a reference to the member function creates  a const reference
     stores it and goes out of scope
 const A& temp = a;
 cout << "Within f(): " << temp.k << "\n";
}

int main() {
   B b(A("hey"));

   cout << "Before f(): " << b.a<< "\n";

   f(b.a);

   cout << "After f(): " << b.a.k << "\n";

   return 0;
}

So when I run this code, I get “hey” as the value everytime. Which seems to imply that a local const reference does not bind itself through life with a passed in member object. Why doesn’t it?

Answer 1

b.a is not a temporary so its lifetime is not affected by any references that are subsequently bound to it.